Proof 30 divides ab(a^2 -b^2)(a^2 +b^2)

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In summary, the assignment was to prove that for all integers a and b, ab(a^2 -b^2)(a^2 +b^2) is divisible by 30. It was noted that this fails if a=b, but the focus was on proving it for a≠b. One approach was to reduce the original expression to (a^5)b-a(b^5) and show that it works, but this was not a helpful step. It was suggested to consider divisibility by 2, 3, and 5, using the fact that they are relatively prime. It was also mentioned that there is an exhaustive way to check this by testing all pairs of values of a and b mod 30.
  • #1
eaglemath15
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I was given the assignment
"Show that for all integers a and b, ab(a^2 -b^2)(a^2 +b^2) is divisible by 30."

Now, I am aware that this fails if a=b, but I want to try to prove this if we assume a≠b.

I've reduced it to the equation (a^5)b-a(b^5), which works, and I know that I could pick the smallest two integers (1 and 2) and show that, but I don't know how to show that it exists for every integer. Should I do proof by contradiction?
 
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  • #2
eaglemath15 said:
I was given the assignment
"Show that for all integers a and b, ab(a^2 -b^2)(a^2 +b^2) is divisible by 30."

Now, I am aware that this fails if a=b,
No, if a = b, the expression above is 0, which is divisible by 30. Note that 0/30 = 0.
eaglemath15 said:
but I want to try to prove this if we assume a≠b.

I've reduced it to the equation (a^5)b-a(b^5), which works
a5b - ab5 is NOT an equation.

What do you mean by "works"? All you have done is to write the original product of three factors as the difference of two terms.

Clearly ab(a2 - b2)(a2 + b2) = a5b - ab5, but I don't see how this is a step in the forward direction.
eaglemath15 said:
, and I know that I could pick the smallest two integers (1 and 2) and show that, but I don't know how to show that it exists for every integer. Should I do proof by contradiction?

For an expression to be divisible by 30, it has to be divisible by 2, 3, and 5.
 
  • #3
Yea, reducing it to [itex]a^{5}b-ab^{5}[/itex] is actually unhelpful as far as I can see, given how I proved it, as it would make things more complicated.. You want to do as Mark said and consider whether or not it is divisible by 2, 3, and 5. If it is divisible by anyone of them, you don't have to worry about the factors you used affecting divisibility by the others since 2,3,5 are prime and therefore relatively prime. So just show that each of those numbers must divide the original equation and you are done.
 
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  • #4
UNChaneul said:
Yea, reducing it to [itex]a^{5}b-ab^{5}[/itex] is actually unhelpful as far as I can see, given how I proved it, as it would make things more complicated.. You want to do as Mark said and consider whether or not it is divisible by 2, 3, and 5. If it is divisible by anyone of them, you don't have to worry about the factors you used affecting divisibility by the others since 2,3,5 are prime and therefore relatively prime. So just show that each of those numbers must divide the original equation and you are done.

Right. Actually the more you factor it the easier it is. There an exhaustive way to check this if you aren't feeling particularly inspired. For example, to check if it's divisible by 5, check all pairs of values of a and b mod 5. There are 25 pairs. Using a form of symmetry will cut that in half. a=b mod 5 obviously works. Now there aren't that many pairs left. The mod 3 and mod 2 cases are much easier.
 
  • #5
Dick said:
Right. Actually the more you factor it the easier it is. There an exhaustive way to check this if you aren't feeling particularly inspired. For example, to check if it's divisible by 5, check all pairs of values of a and b mod 5. There are 25 pairs. Using a form of symmetry will cut that in half. a=b mod 5 obviously works. Now there aren't that many pairs left. The mod 3 and mod 2 cases are much easier.

The reason I had done this was not with respect to the proof, but because it made it easier for me to test whether or not this is actually divisible by 30 (my professor wants us to be in the habit of testing things and not just taking his word that they are true). Thank you for your help though.
 
  • #6
eaglemath15 said:
The reason I had done this was not with respect to the proof, but because it made it easier for me to test whether or not this is actually divisible by 30 (my professor wants us to be in the habit of testing things and not just taking his word that they are true). Thank you for your help though.

Ah I see, good 'ol testing things (but hopefully not "proving" them via American Induction, as my math prof used to joke about).
 
  • #7
eaglemath15 said:
The reason I had done this was not with respect to the proof, but because it made it easier for me to test whether or not this is actually divisible by 30 (my professor wants us to be in the habit of testing things and not just taking his word that they are true). Thank you for your help though.

An exhaustive test of a finite number of cases is a rigorous proof. You could have done this by doing the whole problem that way. But mod 30 you start with 900 cases. The factorization 30=2*3*5 makes it much easier.
 

What does it mean for 30 to divide ab(a^2 -b^2)(a^2 +b^2)?

When we say that 30 divides ab(a^2 -b^2)(a^2 +b^2), it means that 30 is a factor of this expression and the result is a whole number. In other words, when we divide ab(a^2 -b^2)(a^2 +b^2) by 30, we get a whole number without any remainder.

What is the significance of 30 in this expression?

The number 30 is significant because it is a multiple of both 2 and 3, which are prime numbers. This means that 30 has more factors than just 1 and itself, making it easier for it to divide other numbers. In this expression, the presence of 30 ensures that the result will be a whole number and not a decimal or fraction.

How do you prove that 30 divides ab(a^2 -b^2)(a^2 +b^2)?

There are a few ways to prove this. One way is to use the prime factorization of 30 (2 x 3 x 5) and show that each of these prime factors divides the expression without any remainder. Another way is to use the distributive property and simplify the expression to show that 30 is a common factor in each term. You can also use mathematical induction to prove this statement for all values of a and b.

Can you give an example to illustrate this concept?

Sure, let's say a = 4 and b = 2. Then ab(a^2 -b^2)(a^2 +b^2) = 4 x 2(4^2 - 2^2)(4^2 + 2^2) = 4 x 2(16 - 4)(16 + 4) = 4 x 2(12)(20) = 3840. Since 3840 is divisible by 30 (128 x 30 = 3840), we can say that 30 divides ab(a^2 -b^2)(a^2 +b^2).

How is this concept related to divisibility rules?

This concept is related to divisibility rules in that it helps us determine if a number is divisible by another number without actually performing the division. In this case, knowing that 30 divides ab(a^2 -b^2)(a^2 +b^2) allows us to quickly determine if a number is divisible by 30 by checking if it meets the criteria of this expression.

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